# Properties of (the Inverse of) the Gradient of a (Strictly) Convex $$\mathrm{C}^{1}$$ Functions

In this short note, I present the characterizations of convex, continuously differentiable functions in a Eucledean space $$\mathbb{R}^{n}$$ as the functions whose derivatives are monotone. If it is strictly convex, then its derivatives are also strictly monotone and are one-to-one mappings. This note is the supplemental material of Section 3.3 of the paper [5] since it is obvious by the results that if a strictly convex $$\mathrm{C}^{1}$$ function $$S|_{\mathsf{int}}:\mathcal{U}_{\mathsf{int}}\to\mathbb{R}$$ with an $$m$$-dim. manifold $$\mathcal{U}_{\mathsf{int}}\subseteq\mathbb{R}^{m}$$ satisfies $$\mathrm{Im}(\nabla S|_{\mathsf{int}}^{\mathsf{T}})=\mathbb{R}^{m}$$, then

1. 1.

its gradient $$\nabla S|_{\mathsf{int}}$$ is strictly monotone and bijective (by Theorem 1 below);

2. 2.

its inverse $$\sigma\doteq\big{(}\nabla S|_{\mathsf{int}}^{\mathsf{T}}\big{)}^{-1}$$ exists and is also strictly monotone, continuous, and bijective (by Theorem 3 below).

Throughout the note, we let $$D$$ be a given open subset of $$\mathbb{R}^{m}$$.

## (Strictly) Convex $$\mathrm{C}^{1}$$ Functions

As a first step, define convex sets and convex functions as follows:

Definition 1. $$D$$ is convex if:

$$x,y\in D\quad\Longrightarrow\quad tx+(1-t)y\in D\;\,\forall t\in[0,1].\nonumber \\$$

Definition 2. A function $$f:D\to\mathbb{R}$$ is convex if

1. 1)

$$D$$ is convex;

2. 2)

for any $$x,y\in D$$: $$\quad f(tx+(1-t)y)\leq t\,f(x)+(1-t)\,f(y)\qquad\forall t\in[0,1]$$.

$$\;\,f$$ is said to be strictly convex if 2) is replaced by

1. $$\;\,$$2$${}^{\prime}$$)

for any $$x,y\in D$$ such that $$x\neq y$$, $$\quad f(tx+(1-t)y)<t\,f(x)+(1-t)\,f(y)\qquad\forall t\in(0,1)$$.

By a $$\mathrm{C}^{1}$$ function, it is meant a function $$f:D\to\mathbb{R}$$ whose first partial derivatives exist and all continuous. The following lemma is essential in the proof of the theorems.

Lemma 1. Suppose $$f:D\to\mathbb{R}$$ is convex and $$\mathrm{C}^{1}$$. Then, for any $$x$$, $$y\in D$$,

$$f(y)\geq f(x)+\nabla f(x)(y-x).\nonumber \\$$

Moreover, if $$f$$ is strictly convex, then for any $$x$$, $$y\in D$$ such that $$x\neq y$$,

$$f(y)>f(x)+\nabla f(x)(y-x).\nonumber \\$$

Here, we show that the gradient of a convex (resp. strictly convex) $$\mathrm{C}^{1}$$ function $$f:D\to\mathbb{R}$$ is monotone (resp. strictly monotone). For this, we precisely define the monotone function as follows.

Definition 3. A map $$F:D\to\mathbb{R}^{m}$$ on a convex open subset $$D$$ of $$\mathbb{R}^{m}$$ is monotone if for all $$x$$, $$y\in D$$,

$$(y-x)^{\mathsf{T}}(F(y)-F(x))\geq 0.\nonumber \\$$

$$F$$ is said to be strictly monotone if for all $$x$$, $$y\in D$$ such that $$x\neq y$$,

$$(y-x)^{\mathsf{T}}(F(y)-F(x))>0.\nonumber \\$$

Theorem 1. Let $$f:D\to\mathbb{R}$$ be $$\mathrm{C}^{1}$$. Then, its gradient $$\nabla f^{\mathsf{T}}:D\to\mathbb{R}^{m}$$ is monotone (resp. strictly monotone) if $$f$$ is convex (resp. strictly convex).

Proof. For any $$x$$, $$y\in D$$, we have

$$f(y)\geq f(x)+\nabla f(x)(y-x)\nonumber \\$$

by Lemma 1. By choosing the points reverse, we also have

$$f(x)\geq f(y)+\nabla f(y)(x-y)\nonumber \\$$

Adding these inequalities and rearranging it finally yield

$$(\nabla f(y)-\nabla f(x))(y-x)\geq 0.\nonumber \\$$

The proof of the strictly convex case can be done in the exactly same manner with the strict inequality ‘$$>$$’ rather than ‘$$\geq$$.’ ∎

## Properties of the Gradient and its Inverse

When $$f:D\to\mathbb{R}$$ is strictly convex and $$\mathrm{C}^{1}$$ function, then its gradient $$\nabla f^{\mathsf{T}}:D\to\mathbb{R}^{m}$$ is an injective mapping. This can be shown via applying the following lemma:

Lemma 2. A map $$F:D\to\mathbb{R}^{m}$$ is injective if it is strictly monotone.

Proof. The proof can be easily done by contradiction. Suppose there exist $$x$$, $$y\in D$$ such that $$F(x)=F(y)$$ but $$x\neq y$$. Then, we can directly see the contradiction from the definition of the strict monotonicity as

$$0=(y-x)^{\mathsf{T}}\mathbf{0}=(y-x)^{\mathsf{T}}(F(y)-F(x))>0,\nonumber \\$$

which implies “$$0>0$$,” where $$\mathbf{0}$$ is the zero vector in $$\mathbb{R}^{m}$$. Hence, if $$x$$, $$y\in D$$ satisfies $$F(x)=F(y)$$, then $$x=y$$, so that $$F$$ is injective. ∎

Theorem 2. Suppose $$f:D\to\mathbb{R}$$ is strictly convex and $$\mathrm{C}^{1}$$. Then, its gradient $$\nabla f^{\mathsf{T}}$$ is an injective mapping.

Proof. By Theorem 1, $$\nabla f^{\mathsf{T}}$$ is strictly monotone, and hence it is injective by Lemma 2. ∎

The key point given by Theorem 2 is that if $$\mathrm{Im}(\nabla f^{\mathsf{T}})=\mathbb{R}^{m}$$, then $$\nabla f$$ is now bijective, so that the inverse function $$(\nabla f^{\mathsf{T}})^{-1}:\mathbb{R}^{m}\to D$$ is defined over $$\mathbb{R}^{m}$$. This inverse function has the following properties.

Theorem 3. Suppose $$f:D\to\mathbb{R}$$ is strictly convex and $$\mathrm{C}^{1}$$. If $$\mathrm{Im}(\nabla f^{\mathsf{T}})=\mathbb{R}^{m}$$, then

1. 1.

$$\nabla f$$ is bijective and the inverse function $$(\nabla f^{\mathsf{T}})^{-1}:\mathbb{R}^{m}\to\mathcal{D}$$ is defined over $$\mathbb{R}^{m}$$;

2. 2.

the inverse $$(\nabla f^{\mathsf{T}})^{-1}$$ is also strictly monotone, bijective, and continuous.

Proof. The first argument and the fact that the inverse $$(\nabla f^{\mathsf{T}})^{-1}$$ is also bijective are obvious by the above discussion. To show strict monotonicity, suppose $$z,w\in\mathbb{R}^{m}$$ and let $$x^{\mathsf{T}}\doteq(\nabla f)^{-1}(z)$$ and $$y^{\mathsf{T}}\doteq(\nabla f)^{-1}(w)$$. Then, since $$\nabla f$$ (and $$(\nabla f)^{-1}$$) is bijective, we have $$z=\nabla f^{\mathsf{T}}(x)$$ and $$w=\nabla f^{\mathsf{T}}(y)$$. Therefore, by strict monotonicity of $$\nabla f$$ (Theorem 1), we conclude that

$$\big{(}(\nabla f)^{-1}(z)-(\nabla f)^{-1}(w)\big{)}(z-w)=(x-y)^{\mathsf{T}}(\nabla f^{\mathsf{T}}(x)-\nabla f^{\mathsf{T}}(y))>0,\nonumber \\$$

meaning that $$(\nabla f^{\mathsf{T}})^{-1}$$ is also strictly monotone. Finally, as $$\nabla f^{\mathsf{T}}$$ is bijective and continuous, the application of invariance of domain [4] proves continuity of $$(\nabla f^{\mathsf{T}})^{-1}$$.     ∎